Unit 1-4 Review Calculations
Let A and B be two disjoint events such that P(A) = 0.1 and P(B) = 0.47. What is P(A or B)?
0.57 add
Let A and B be two independent events such that P(A) = 0.33 and P(B) = 0.57. What is P(A or B)?
0.7119 add; multiply, subtract the two
After she rolls it 37 times, Joan finds that she's rolled the number 2 a total of six times. What is the empirical probability that Joan rolls a 2?
16.22% 6/37
The distribution of the amount of money spent by students on textbooks in a semester is approximately normal in shape with a mean of 424 and a standard deviation of 25. According to the standard deviation rule, approximately 95% of the students spent between $___ and $ ____ on textbooks in a semester.
374 and 474 424-2*(25) and 424+2*(25)
A recent survey asks 89 students, How many hours do you spend on the computer in a typical day? Of the 89 respondents, 2 said 1 hour, 4 said 2 hours, 14 said 3 hours, 22 said 4 hours, 18 said 5 hours, 15 said 6 hours, 7 said 7 hours, 5 said 8 hours, 2 said 9 hours. What is the average (mean) number of hours spent on the computer?
4.8 =SUMPRODUCT(# of students, # of hours)/SUM(# of students)
What percentage of students earned a grade of less than 90?
88% less than 90: [80-90) column: 15/17*100
The distribution of IQ (Intelligence Quotient) is approximately normal in shape with a mean of 100 and a standard deviation of 13. According to the standard deviation rule, ____ % of people have an IQ between 61 and 139. Do not round.
99.7% 100-61 = 39 139-100 = 39 39/13 = 3 : 3 falls between 99.7%
Suppose that P(A) = 0.96. Which of the following is the best interpretation of this statement?
Event A is extremely likely, but in a long sequence of trials, it occasionally will not occur.
True or false? The researcher could produce a narrower confidence interval by increasing the sample size to 150.
True
Gardeners on the west coast of the United States are investigating the difference in survival rates of two flowering plants in drought climates. Plant A has a survival rate of 0.67 and plant B has a survival rate of 0.47. The standard error of the difference in proportions is 0.093. What is the margin of error for a 99% confidence interval? Use critical value z = 2.576. MOE =
0.240 =(2.576)*(0.093)
The distribution of IQ (Intelligence Quotient) is approximately normal in shape with a mean of 100 and a standard deviation of 20. According to the standard deviation rule, only ____ % of people have an IQ over 140.
2.5% 140-100 = 40 40/20 = 2 : 2 falls between 2.5% (1 falls between 16%)
Based on the table above, the percentage of Married participants, who were Obese is:
34.6% =146/422*100 =Married & Obese column/Married total
Calculate the standard error for a survey comparing proportions of cognition-enhancing drug use of fraternity members to non-fraternity members, where p1 = 0.28, n1 = 146, p2 = 0.22, n2 = 95. Round all calculations to the thousandth decimal place.
=(0.28*(1-0.28)/146) for p1 =(0.22*(1-0.22)/95) for p2 =SQRT(p1 + p2) = 0.056
The faculty senate at a large university wanted to know what proportion of the students thought foreign language classes should be required for everyone. The statistics department offered to cooperate in conducting a survey, and a simple random sample of 500 students was selected from all the students enrolled in statistics classes. A survey form was sent by email to these 500 students. In this case, which of the following is the population of interest?
All students at the university
True or false? Finding a random sample with a mean this low in a population with mean 7 and standard deviation 2 is very unlikely.
False
A politician claims that a larger proportion of members of the news media are Democrats when compared to the general public. Let p1 represent the proportion of the news media that is Democrat and p2 represent the proportion of the public that is Democrat. What are the appropriate null and alternative hypotheses that correspond to this claim?
H0: p1 - p2 = 0; Ha: p1 - p2 > 0
The outlier on the graph is likely due to an error in recording the data. Which of the following statements is true?
If the outlier were removed, the correlation coefficient (r) would increase.
In 2011, the Institute of Medicine (IOM), a non-profit group affiliated with the US National Academy of Sciences, reviewed a study measuring bone quality and levels of vitamin-D in a random sample from bodies of 675 people who died in good health. 8.5% of the 82 bodies with low vitamin-D levels (below 50 nmol/L) had weak bones. Comparatively, 1% of the 593 bodies with regular vitamin-D levels had weak bones.
No, there are not at least 10 people with weak bones and 10 people with strong bones in each group.
The sample suggests that the food is safe, but it actually is not safe.
Type II
In this experiment researchers randomly assigned smokers to treatments. Of the 164 smokers taking a placebo, 22 stopped smoking by the 8th day. Of the 275 smokers taking only the antidepressant buproprion, 81 stopped smoking by the 8th day. Calculate the estimated standard error for the sampling distribution of differences in sample proportions. The estimated standard error =
p1: 22/164 n1: 164 =(p1*(1-p1)/164 for p1 p2: 81/275 n2: 275 =(p2*(1-p2)/275 for p2 =SQRT(p1 + p2) = 0.038
The distribution of the amount of money spent by students on textbooks in a semester is approximately normal in shape with a mean of: μ= 409 and a standard deviation of: σ= 24. According to the standard deviation rule, almost 16% of the students spent more than what amount of money on textbooks in a semester?
$433 409+24
The distribution of scores on a recent test closely followed a Normal Distribution with a mean of 22 points and a standard deviation of 2 points. For this question, DO NOT apply the standard deviation rule. (a) What proportion of the students scored at least 28 points on this test, rounded to five decimal places? (b) What is the 85 percentile of the distribution of test scores, rounded to three decimal places?
(a) =1-NORM.DIST(28,22,2,1) (b) =NORM.INV(0.85,22,2)
A study was made of seat belt use among children who were involved in car crashes that caused them to be hospitalized. It was found that children not wearing any restraints had hospital stays with a mean of 7.37 days and a standard deviation of 2.75 days with an approximately normal distribution. (a) Find the probability that their hospital stay is from 5 to 6 days, rounded to five decimal places. (b) Find the probability that their hospital stay is greater than 6 days, rounded to five decimal places.
(a) =NORM.DIST(6,7.37,2.75,1)-NORM.DIST(5,7.37,2.75,1) (b) =1-NORM.DIST(6,7.37,2.75,1)
Let A and B be two disjoint events such that P(A) = 0.57 and P(B) = 0.2. What is P(A and B)?
0
Let A and B be two independent events such that P(A) = 0.3 and P(B) = 0.6. What is P(A and B)?
0.18 multiply
An urn contains 12 red marbles, 23 blue marbles, and 44 yellow marbles. One marble is to be chosen from the urn without looking. What is the probability of choosing a red or a blue marble?
0.4430 12+23+44 = 79 12/79 + 23/79
The heights of students at a college are normally distributed with a mean of 175 cm and a standard deviation of 6 cm. One might expect in a sample of 1000 students that the number of students with heights less than 163 cm is:
23 =NORM.DIST(163,175,6,1)*1000
The p-value is 0.06. If we conduct this test at a 5% level of significance, what would be an appropriate conclusion?
Fail to Reject H0 , do not support Ha .
When the student summarizes the data, she finds that 42 of the 50 business students and 38 of the 70 nursing students admitted to cheating in their courses. True or false? The counts suggest that the normal model is a good fit for the sampling distribution of sample differences.
False
Researchers conduct a hypothesis test to determine if the proportion of U.S. residents consuming recommended levels of calcium is different among women and men. The p-value is 0.035, and researchers conduct this test at a 5% level of significance. What does a p-value of 0.035 mean?
If calcium consumption is the same for women and men, there is a 3.5% chance that future studies will show differences in calcium consumption greater than observed in this study.
The difference between teenage female and male depression rates estimated from two samples is 0.07. The estimated standard error of the sampling distribution is 0.04. What is the 95% confidence interval? Use the critical value z = 1.96.
Interval: =(0.07 - 1.96 *times 0.04), =(0.07 + 1.96 *times 0.04) (-0.01, 0.15)
Which class has greater variability in students' estimate of the number of marshmallows?
Ms. Apple's class
Which class has a greater percentage of estimates between 50 and 100 marshmallows?
Ms. Banana's class
To test the claim (at 5% significance) that the proportion of U.S. residents who consume recommended levels of vitamin A is higher among women than men, researchers set up the following hypotheses: In this hypothesis test which of the following errors is a Type I error?
Researchers conclude that a larger proportion of women consume the recommended daily intake of vitamin A when there is actually no difference between vitamin A consumption for women and men.
The histogram below displays the distribution of 50 ages at death due to trauma (accidents and homicides) that were observed in a certain hospital during a week. Which of the following best describes the shape of the histogram?
Right-skewed with a possible outlier
Which of the following variables is discrete? Check all that apply.
Shoe size Dress size
The administration at GSU wants to estimate the number of parking spaces they will need next year. They survey 80 students; 75 of the students in the sample drive to campus by themselves each day
The sample needs to be random but we don't know if it is. The actual count of those who do not drive to campus is too small. n(1−^p) is not greater than 10.
Based on the limited amount of available student parking spaces on the GSU campus, students are being encouraged to ride their bikes (when appropriate)
The sample needs to be random but we don't know if it is. The actual count of bike riders is too small. n*^p is not greater than 10.
GSU'sRialto Center for the Performing Arts wanted to investigate why ticket sales for the upcoming season significantly decreased from last year's sales. The marketing staff collected data from a survey of community residents. Out of the 110 people surveyed, only 7 received the concert brochure in the mail.
The sample needs to be random, but we don't know if it is. The actual count of community residents who received the concert brochure by mail is too small. n^p is not greater than 10.
Suppose that the correlation r between two quantitative variables was found to be r=0. Which of the following is the best interpretation of this correlation value?
There is no linear relationship between the two variables.
In the population, 8% of males have had a kidney stone. Suppose a medical researcher randomly selects two males from a large population. Let A represent the event "the first male has had a kidney stone." Let B represent the event "the second male has had a kidney stone." True or false? A and B are independent events.
True
The city council hired three college interns to measure public support for a large parks and recreation initiative in their city. The interns mailed surveys to 500 randomly selected participants in the current public recreation program. They received 150 responses. True or false? Even though the sample is random, it is NOT representative of the population of interest.
True
We know that narrower confidence intervals give us a more precise estimate of the true population proportion. Which of the following could we do to produce higher precision in our estimates of the population proportion?
We can select a lower confidence level and increase the sample size.
In which of the following scenarios would the distribution of the sample mean x-bar be normally distributed? Check all that apply.
We take repeated random samples of size 15 from a population that is normally distributed. We take repeated random samples of size 50 from a population of unknown shape.
A researcher wants to determine if preschool attendance is associated with high school graduation for low-income students. She randomly assigns low-income children to two groups; one group will attend preschool program, the second group will not attend preschool. The researcher plans to follow the children in the study for 20 years and observe whether or not they graduate from high school. Which of the following is the response variable in this study?
Whether or not a subject graduates high school
Commute times in the U.S. are heavily skewed to the right. We select a random sample of 240 people from the 2000 U.S. Census who reported a non-zero commute time. In this sample the mean commute time is 28.9 minutes with a standard deviation of 19.0 minutes. Can we conclude from this data that the mean commute time in the U.S. is less than half an hour? Conduct a hypothesis test at the 5% level of significance. What is the p-value for this hypothesis test?
se = 19/SQRT(240) t value= (28.9-30)/se p value =T.DIST(t value, 240-1, 1) = 0.1853
Find the p-value for the hypothesis test. A random sample of size 50 is taken. The sample has a mean of 420 and a standard deviation of 81. H0: µ = 400 Ha: µ > 400 The p-value for the hypothesis test is
se = 81/SQRT(50) t value= (420-400)/se p value =T.DIST.RT(t value, 50-1) = 0.0435
Suppose that we conduct a hypothesis test in which a Type II error is very serious. But the Type I error is not very serious. Which level of significance is the best choice?
α = 0.05
A group of engineers developed a new design for a steel cable. They need to estimate the amount of weight the cable can hold. The weight limit will be reported on cable packaging. The engineers take a random sample of 46 cables and apply weights to each of them until they break. The 46 cables have a mean breaking weight of 779.2 lb. The standard deviation of the breaking weight for the sample is 15.3 lb. Find the 90% confidence interval to estimate the mean breaking weight for this type cable.
=T.INV(0.95, 46-1) Interval: =779.2- value above* (15.3/SQRT(46)) , =779.2 + value above *(15.3/SQRT(46)) (775.41, 782.99)
A florist determines the probabilities for the number of flower arrangements they deliver each day. x 19 20 21 22 23 P(x) 0.21 0.23 0.30 0.14 0.12 Find the mean, variance, and standard deviation of the distribution rounded to 4 decimal places. Approximately how many arrangements should the florist expect to deliver each week, rounded to the nearest whole number?
Mean = 20.73 =SUMPRODUCT(x, P(x)) Variance = 1.6171 =SUMPRODUCT(x - mean)^2, P(x)) Standard deviation = 1.2717 =SQRT(variance) 145 =Mean * 7
Calculate the row percentages for participants who were Obese .
Perc 1 Dating= 24.9% =83/334*100 (total Obese) Perc 2 Cohabiting= 30.8% =103/334*100 Perc 3 Married= 44.3% =148/334*100 Total should be 100%
Which of the following is the best description of the relationship between X and Y as it appears in the scatterplot?
Positive linear relationship with outlier(s)
In 2015 as part of the General Social Survey, 1194 randomly selected American adults responded to this question: Of the respondents, 502 replied that America is doing about the right amount. What is the 95% confidence interval for the proportion of all American adults who feel that America is doing about the right amount to protect the environment.
p hat = 502/1194 se = SQRT(p hat*(1-p hat)/sample size) Interval: (p hat - 1.96 *(se) , p hat + 1.96 *(se) (0.392, 0.448)
A group of 54 college students from a certain liberal arts college were randomly sampled and asked about the number of alcoholic drinks they have in a typical week. The purpose of this study was to compare the drinking habits of the students at the college to the drinking habits of college students in general. In particular, the dean of students, who initiated this study, would like to check whether the mean number of alcoholic drinks that students at his college in a typical week differs from the mean of U.S. college students in general, which is estimated to be 4.73. The group of 54 students in the study reported an average of 5.25 drinks per with a standard deviation of 3.98 drinks. Find the p-value for the hypothesis test.
t value =3.98/SQRT(54) 5.25-4.73 = 0.52 0.52/t value = 0.906102 p value: =T.DIST.2T(0.960102 ,54 - 1) = 0.3414
According to the information that comes with a certain prescription drug, when taking this drug, there is a 17% chance of experiencing nausea (N) and a 43% chance of experiencing decreased sexual drive (D). The information also states that there is a 10% chance of experiencing both side effects. What is the probability of experiencing only nausea?
0.07 n not n total fill in chart: only nausea is top middle: n, not d
Dogs are inbred for such desirable characteristics as blue eye color, but an unfortunate by-product of such inbreeding can be the emergence of characteristics such as deafness. A 1992 study of Dalmatians (by Strain and others, as reported in The Dalmatians Dilemma) found the following: (i) 31% of all Dalmatians have blue eyes. (ii) 38% of all Dalmatians are deaf. (iii) If a Dalmatian has blue eyes, there is a 42% chance that it is deaf. What is the probability that a randomly chosen Dalmatian is blue-eyed and deaf?
0.31 * 0.42 = 0.1302
According to the information that comes with a certain prescription drug, when taking this drug, there is a 17% chance of experiencing nausea (N) and a 50% chance of experiencing decreased sexual drive (D). The information also states that there is a 11% chance of experiencing both side effects. What is the probability of experiencing neither of the side effects?
0.44 n d not d total not n total fill in chart: neither is middle: not n, not d
The random variable X, representing the number of accidents in a certain intersection in a week, has the following probability distribution: x 0 1 2 3 4 5 P(X = x) 0.20 0.30 0.20 0.15 0.10 0.05 On average, how many accidents are there in the intersection in a week?
1.8 =SUMPRODUCT(x, P(X = x) )
Suppose the time to complete a 200-meter backstroke swim for female competitive swimmers is normally distributed with a mean μ = 141 seconds and a standard deviation σ = 7 seconds. What is the completion time for the 200-meter backstroke for a female with a z-score of −1.64? (Round answer to 1 decimal place.)
129.5 141-1.64*7
A certain medical test is known to detect 65% of the people who are afflicted with the disease Y. If 10 people with the disease are administered the test, what is the probability that the test will show that: All 10 have the disease, rounded to four decimal places? At least 8 have the disease, rounded to four decimal places? At most 4 have the disease, rounded to four decimal places?
All 10: =BINOM.DIST(10,10,0.65,0) At least 8: =1-BINOM.DIST(7,10,0.65,1) At most 4: =BINOM.DIST(4,10,0.65,1)
Which of the following scenarios are Binomial?
An engineer chooses a SRS of 10 switches from a shipment of 10,000 switches. Suppose 10% of the switches in the shipment are bad. The engineer counts the number X of bad switches in the sample. You observe the sex of the next 20 children born at a local hospital: X is the number of girls among them.
Suppose that the handedness of the last fifteen U.S. presidents is as follows: 40% were left-handed (L) 47% were Democrats (D) If a president is left-handed, there is a 13% chance that the president is a Democrat. Based on this information on the last fifteen U.S. presidents, is "being left-handed" independent of "being a Democrat"?
No, since 0.47 is not equal to 0.13
A student survey was conducted at a major university, and data were collected from a random sample of 750 undergraduate students. One variable that was recorded for each student was the student's answer to the question "With whom do you find it easiest to make friends? Opposite sex/same sex/no difference." These data would be best displayed using which of the following?
Pie chart
Which of the following is an example of stratified sampling?
Proponents of a local ballot measure conduct a survey of the city by randomly selecting 100 potential voters from each of its 18 zip codes.
If we set up our null and alternative hypotheses as follows: H0:p=0.37 Ha:p>0.37 and find that: "p-value"=0.418. Does this provide enough evidence to support Jason's claim? Use an α=0.05 level of significance.
Since the p-value > α, do not reject the null hypothesis.
A factory produces plate glass with a mean thickness of 4mm and a standard deviation of 1.1mm. A simple random sample of 100 sheets of glass is to be measured, and the mean thickness of the 100 sheets is to be computed. What is the probability that the average thickness of the 100 sheets is less than 3.91 mm?
Standard Deviation =1.1/SQRT(100) =NORM.DIST(3.91,4,Standard Deviation,1)
Suppose that the distribution for total amounts spent by students vacationing for a week in Florida is normally distributed with a mean of 650 and a standard deviation of 120. Suppose you take a simple random sample (SRS) of 15 students from this distribution. What is the probability that a SRS of 15 students will spend an average of between 600 and 700 dollars?
Standard Deviation =120/SQRT(15) =NORM.DIST(700,650,Standard Deviation,1)-NORM.DIST(600,650,Standard Deviation,1)
Suppose that 88% of all dialysis patients will survive for at least 5 years. In a simple random sample of 100 new dialysis patients, what is the probability that the proportion surviving for at least five years will exceed 80%, rounded to 5 decimal places?
Standard Deviation =SQRT(0.88*(1-0.88)/100) =1-NORM.DIST(0.80,0.88,Standard Deviation,1)
According to a 2014 research study of national student engagement in the U.S., the average college student spends 17 hours per week studying. A professor believes that students at her college study less than 17 hours per week. The professor distributes a survey to a random sample of 80 students enrolled at the college. From her survey data the professor calculates that the mean number of hours per week spent studying for her sample is: ¯x= 15.6 hours per week with a standard deviation of s = 4.5 hours per week. The professor chooses a 5% level of significance. What can she conclude from her data?
The data supports the professor's claim. The average number of hours per week spent studying for students at her college is less than 17 hours per week.
Suppose we take repeated random samples of size 20 from a population with a mean of 60 and a standard deviation of 8. Which of the following statements is true about the sampling distribution of the sample mean (x̄)? Check all that apply.
The distribution will be normal as long as the population distribution is normal. The distribution's mean is the same as the population mean 60.
Of the 1251 patients who were mailed surveys, 250 patients responded. For various reasons, researchers used only 225 of the completed surveys. 31 out of 116 female cancer patients reported being past smokers, and 57 out of 109 male cancer patients reported being past smokers. Calculate the difference between the corresponding sample proportions ^p1−^p2 (female minus male). Round the answer to 4 decimal places.
female p1: 31/116 male p2: 57/109 p1-p2 = -0.2557
The SAT is the most widely used college admission exam. (Most community colleges do not require students to take this exam.) The mean SAT math score varies by state and by year, so the value of µ depends on the state and the year. But let's assume that the shape and spread of the distribution of individual SAT math scores in each state is the same each year. More specifically, assume that individual SAT math scores consistently have a normal distribution with a standard deviation of 100. An educational researcher wants to estimate the mean SAT math score (μ) for his state this year. The researcher chooses a random sample of 661 exams in his state. The sample mean for the test is 481. Find the 90% confidence interval to estimate the mean SAT math score in this state for this year. (Note: The critical z-value to use, zc, is: 1.645.)
mean = 481 se = 100/SQRT(661) Interval: =481 - 1.645*(se) , 481 + 1.645*(se) (474.602, 487.398)
A survey asks a random sample of 1500 adults in Ohio if they support an increase in the state sales tax from 5% to 6%, with the additional revenue going to education. Let ^p denote the proportion in the sample who say they support the increase. Suppose that 79% of all adults in Ohio support the increase. The standard deviation of the sampling distribution is
p hat = 0.79 =SQRT(p hat*(1- p hat)/1500) = 0.0105
According to a Pew Research Center study, in May 2011, 35% of all American adults had a smart phone (one which the user can use to read email and surf the Internet). A communications professor at a university believes this percentage is higher among community college students. She selects 386 community college students at random and finds that 153 of them have a smart phone. Then in testing the hypotheses: H0: p = 0.35 versus Ha: p > 0.35, what is the test statistic?
p hat = 153/386 se =SQRT(0.35*(1-0.35)/sample size) test statistic =(p hat - 0.35)/se
Suppose we take a poll (random sample) of 3557 students classified as Juniors and find that 3298 of them believe that they will find a job immediately after graduation. What is the 99% confidence interval for the proportion of GSU Juniors who believe that they will, immediately, be employed after graduation.
p hat = 3298/3557 se =SQRT(p hat*(1-p hat)/sample size) Interval: p hat - 2.56*(se) , p hat + 2.56*(se) (0.916, 0.938)
